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gtnhenbr [62]
3 years ago
6

A student sees a newspaper ad for an apartment that has 1330 square feet (ft^2) or floor space. How many square meters of area a

re there?
Physics
1 answer:
Natasha2012 [34]3 years ago
6 0
<span>=1330 ft^2 x (1m^2/3.25 ft^2)
=124 m^2

</span>
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A person is sitting with one leg outstretched and stationary, so that it makes an angle of θ = 27.5° with the horizontal, as the
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here we will use the torque balance about the knee joint

here we can say that

\tau_g = \tau_m

here torque due to weight is given as

\tau_g = 40.1 cos\theta*(l_1 + l_2)

\tau_g = 40.1 cos27.5*(0.105 + 0.150)

\tau_g = 9.07 Nm

now torque due to applied force of muscle

\tau_m = M*sin\alpha * l_1

\tau_m = M*sin30* 0.105

now by torque balance we will have

9.07 = M*0.5*0.105

M = 173 N

so here the magnitude of m will be 173 N

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3 years ago
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What is the voltage of a computer with 30 Ω of resistance and 15 amps of current?
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What is the type of intermolecular force in methylated spirits
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A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

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Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

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